Elements of Geometry and Trigonometry from the Works of A.M. Legendre: Adapted to the Course of Mathematical Instruction in the United States |
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Page 57
... whence , BA = AB ; A B A B = = D C F E AP 22 ; whence , BC = AD ; ; whence , = BE AF ; B H = ; whence , BG = AH ; Α G & c . & c . , Adding and factoring , we have , B ( A + C + E + G + & c . ) = A ( B + D + F + H + & c . ) : hence ...
... whence , BA = AB ; A B A B = = D C F E AP 22 ; whence , BC = AD ; ; whence , = BE AF ; B H = ; whence , BG = AH ; Α G & c . & c . , Adding and factoring , we have , B ( A + C + E + G + & c . ) = A ( B + D + F + H + & c . ) : hence ...
Page 58
... whence , F E 31 72 B D = A H = འཁ -༤ and , E : F :: G : H ; whence , Multiplying the equations , member by member , we have , BF DH = AE CG ; whence , AE : BF :: CG : DH ; which was to be proved . Cor . 1. If the corresponding terms of ...
... whence , F E 31 72 B D = A H = འཁ -༤ and , E : F :: G : H ; whence , Multiplying the equations , member by member , we have , BF DH = AE CG ; whence , AE : BF :: CG : DH ; which was to be proved . Cor . 1. If the corresponding terms of ...
Page 76
... whence , angle ACB : angle ACD DIOB arc AB : arc 40 . Conceive the arc AB to be divided into equal parts , each less than DO : there will be at least one point of division between D and 0 ; let I be that point ; and draw CI . Then the ...
... whence , angle ACB : angle ACD DIOB arc AB : arc 40 . Conceive the arc AB to be divided into equal parts , each less than DO : there will be at least one point of division between D and 0 ; let I be that point ; and draw CI . Then the ...
Page 109
... whence , by addition , recollecting that AE is equal to CE and BE to DE , we have , AB2 + BC2 + CD2 + DA2 ᎠᎪ = 4CE + 4DE2 ; but , 4 CE is equal to AC , and 4DE2 to BD2 ( P. VIII . , C. ) : hence , AB2 + BC2 + CD2 + DA2 = AC2 + BD2 ...
... whence , by addition , recollecting that AE is equal to CE and BE to DE , we have , AB2 + BC2 + CD2 + DA2 ᎠᎪ = 4CE + 4DE2 ; but , 4 CE is equal to AC , and 4DE2 to BD2 ( P. VIII . , C. ) : hence , AB2 + BC2 + CD2 + DA2 = AC2 + BD2 ...
Page 120
... DC ; which was to be proved . and , Cor . 1. From the proportions , BC : AB :: AB : BD , BC : AC : AC : DC , we have ( B. II . , P. I. ) , • and , AB2 = BC × BD , ᎪᏛ = BC DC ; whence , by addition , AB2 + AC2 = BC 120 GEOMETRY .
... DC ; which was to be proved . and , Cor . 1. From the proportions , BC : AB :: AB : BD , BC : AC : AC : DC , we have ( B. II . , P. I. ) , • and , AB2 = BC × BD , ᎪᏛ = BC DC ; whence , by addition , AB2 + AC2 = BC 120 GEOMETRY .
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Common terms and phrases
AB² AC² adjacent angles altitude angle ACB apothem Applying logarithms base and altitude bisect centre chord circle circumference circumscribed coincide cone consequently convex surface corresponding cosec Cosine Cotang cylinder denote diagonals diameter distance divided draw drawn edges equally distant feet find the area Formula frustum given angle given straight line greater hence homologous hypothenuse included angle interior angles intersection less Let ABC log sin lower base mantissa mean proportional measured by half number of sides opposite parallel parallelogram parallelopipedon perimeter perpendicular plane MN polyedral angle polyedron prism PROPOSITION proved pyramid quadrant radii radius rectangle regular polygons right angles right-angled triangle Scholium secant segment semi-circumference side BC similar sine six right slant height sphere spherical polygon spherical triangle square Tang tangent THEOREM triangle ABC triangular prisms triedral angle upper base vertex vertices whence